1. Hello everyone,

I'm new to this forum. Can you clear my doubt about Einstein's special relativity theory?

It says anything can not travel more than the speed of light. because, when velocity reaches the speed of light, mass of that body reaches infinite and it can not accelerate. but, consider engine pushes a rocket, mass of the rocket is always same with respect to engine of that rocket. because, velocity of that rocket is zero relative to engine. so, engine can accelerate rocket further and further at constant rate. Then what about Universe's speed limit?  2.

3. First, the trivial answer. In the frame of the rocket, the rocket's velocity is zero by definition, so it never goes faster than light. We don't even need relativity to get that answer.

Hence, we must choose a frame not attached to the rocket. Let us take the frame where the rocket was not moving before it was started. In this frame, what we are doing is converting the chemical energy and perhaps some of the rest-mass energy of the propellant into the kinetic energy of the moving rocket. The amount of this energy is, of course, finite and does not increase with speed. The kinetic energy of the rocket if it were moving at the speed of light would be infinite. Therefore there isn't enough energy to reach, much less exceed, the speed of light.

In other words, we can't access kinetic energy to further increase the energy of the system. The loss of speed from the use of the kinetic energy exactly balances the gain in kinetic energy obtained, of course.  4. Thank you sir, Originally Posted by mvb The kinetic energy of the rocket if it were moving at the speed of light would be infinite. Therefore there isn't enough energy to reach, much less exceed, the speed of light.
In particle accelerators, particles could not be accelerated more than the speed of light. because, Particle accelerator accelerates particle and particle is travelling at .9999 c speed (with approximately infinite mass and kinetic energy) relative to accelerator. so, accelerator can not accelerate particle more than c (Here the frame is accelerator).
But in a rocket, engine accelerates rocket and if we take the frame with engine, Engine can accelerate more and more. because, rocket has its rest mass relative to engine. Frame is with engine. because engine dose the work. so, engine needs constant energy from fuel for constant acceleration.

Particle relative to accelerator (Frame)
Rocket relative to engine (Freme)  5. Originally Posted by Dinesh balu Hello everyone,

I'm new to this forum. Can you clear my doubt about Einstein's special relativity theory?

It says anything can not travel more than the speed of light. because, when velocity reaches the speed of light, mass of that body reaches infinite and it can not accelerate. but, consider engine pushes a rocket, mass of the rocket is always same with respect to engine of that rocket. because, velocity of that rocket is zero relative to engine. so, engine can accelerate rocket further and further at constant rate. Then what about Universe's speed limit?
There's a couple of ways of approaching this. One is from the original start frame from which the rocket started accelerating. One thing to keep in mind is that according to this frame,as the rocket gets closer and closer to the speed of light, time runs slower for the rocket, it length contracts, etc. The end result is that in order to get the exhaust velocity with respect to the the Rocket as measured by this frame, you have to use the following equation for the relativistic addition of velocities: If v is the velocity of the ship, then u is the velocity of the exhaust as measured by the ship. (in this case, u would be negative because it is opposite in direction from v.)

Then take the difference between S and v to get the exhaust velocity with respect to the ship.

For example, if v= 0.6c and u= 0.5c

then S= 0.143c and the difference between S and v is 0.457c

Notice that this is less than the 0.5 c as measured by the ship.

Since the exhaust velocity determines how much "push" the engines give, we have seen a drop in thrust.

Now consider what happens at v=0.999c

then S= 0.997c and the exhaust velocity with respect to the ship drops to .0001997c

The closer the ship gets to c, the less faster the thrust, and thus its acceleration, falls. And it falls off so quickly, that the rocket, while it can get closer and closer to c, never quite gets there.

The second way is to consider things from the ship's perspective.

To make things easier to follow, we are going to have our ship drop off marker buoys at regular intervals as it travels. The buoys maintain the speed the ship had at the moment they were dropped.

He drops the first buoy when he has reached 0.1c relative to his starting point. He then continues until he reaches 0.1c relative to the first buoy. So how fast is he traveling relative to his starting point?

You have to use the same relativistic velocity addition as above, thought this time U is the velocity of the first buoy to the starting point, and v is the velocity of the ship with respect to the first buoy.

We get an answer of 0.198c, a little less than if we had just added 0.1c and 0.1c.

He drops a second buoy and continues on.

When he is moving at 0.1c with respect to the second buoy, he is moving at 0.198c with respect to the first buoy, and 0.292c with respect to his starting point. (The difference is beginning to grow)

As he keeps dropping buoys and determining how fast he is moving with respect to his starting point, he finds that he is gaining less speed every time.

If you go back to the formula above, it is easy to see that no matter how large v and u are, as long as they are less than c, the final answer will be less than c. And since any rocket must start at less than c, and its change in velocity can be broken up in to "less than c" segments, the total addition of all those segments will always be less than c.

All this leads up to the relativistic rocket equation: Where Ve is the exhaust velocity with as measured by the rocket and MR is the mass ratio (fully fueled rocket divided by rocket after expending fuel.)

As delta v approaches c, MR approaches infinity.  6. MODERATOR NOTE : I am moving this over to Physics, as it is not directly related to Astronomy & Cosmology.  7. Originally Posted by Dinesh balu But in a rocket, engine accelerates rocket and if we take the frame with engine, Engine can accelerate more and more. because, rocket has its rest mass relative to engine. Frame is with engine. because engine dose the work. so, engine needs constant energy from fuel for constant acceleration.
It is common to see explanations based on the infinite energy requirements of accelerating a massive object to the speed of light and beyond, but I actually prefer an explanation that has nothing to do with the energy requirements. You are correct in saying that the rocket can constantly accelerate forever, and over this infinite amount of time, given the constant rate of energy expended, the total energy will be infinite. But because the speed of light is the same for all observers, no matter how long one accelerates, one will never get any closer to the speed of light. Thus, in the frame of reference of the rocket, the rocket constantly accelerates forever, but in an inertial frame of reference, the rocket will asymptotically approach the speed of light. It should be remarked that a constant acceleration in the rocket's frame of reference is not a constant acceleration in the inertial frames of reference.  8. Good day sirs,

Now I am clear with this. well clarified ..! Thank you  9. Originally Posted by Dinesh balu Hello everyone,

I'm new to this forum. Can you clear my doubt about Einstein's special relativity theory?

It says anything can not travel more than the speed of light. because, when velocity reaches the speed of light, mass of that body reaches infinite and it can not accelerate. but, consider engine pushes a rocket, mass of the rocket is always same with respect to engine of that rocket. because, velocity of that rocket is zero relative to engine. so, engine can accelerate rocket further and further at constant rate. Then what about Universe's speed limit?
Here is another explanation (that doesn't use the mass-energy). In relativity, speed is related to acceleration via: When as in your example, with at all times.  10. I think it should be enough to note that in Minkowski spacetime, 4-velocity and 4-acceleration are everywhere orthogonal, and that the Minkowski norm of the 4-velocity is always constant at -1 ( hyperbolic geometry ! ); hence :  If we assume that the acceleration is constant at some value g and parallel along some coordinate axis, we also have There isn't any physically possible combination of a and u that fulfils these three equations in such a way so as to give a linear speed of c, except the trivial a=0, u=c along some given direction, which corresponds to a photon.

In other words, the fact that nothing can accelerate to the speed of light is a simple consequence of the geometry of Minkowski space-time, specifically of the fact that the norm of the 4-velocity is constant, while 4-velocity and 4-acceleration are orthogonal everywhere. We don't need to bring energy into this at all, it is all purely geometric.

Or am I seeing this wrong ?  11. Originally Posted by Markus Hanke the Minkowski norm of the 4-velocity is always constant at -1 ( hyperbolic geometry ! ); hence : Actually, ... ALWAYS ... irrespective of the indefinite nature of the metric or its signature (yes, I've fallen into a similar trap!). This can be seen by going back to the definition of and the metric:  Divide both sides by :   12. It takes energy to make something move faster than the speed it's already traveling at. (Acceleration.)

Anything that's currently going at less than light speed will need infinite energy to increase its speed to that of light.

Hmm, I just had a thought (which is likely to be wrong, but I'm just musing.)

If Higgs bosons permeate spacetime, could an object moving at near light speed gain mass because it's colliding with all these Higgs bosons as it travels, and that the Higgs bosons are slightly "sticky" (imagine they stick to the object for n number of Planck-lengths of time) so the faster something moves, the more bosons it collides with, and therefore there's an imbalance between the number of bosons getting "stuck" to the object for n amount of time than there is number of bosons falling away from the object moving at near-light speed?

I'm not sure I explained my thoughts correctly.  13. Originally Posted by Daecon If Higgs bosons permeate spacetime, could an object moving at near light speed gain mass because it's colliding with all these Higgs bosons as it travels, and that the Higgs bosons are slightly "sticky" (imagine they stick to the object for n number of Planck-lengths of time) so the faster something moves, the more bosons it collides with, and therefore there's an imbalance between the number of bosons getting "stuck" to the object for n amount of time than there is number of bosons falling away from the object moving at near-light speed?
No. For one thing, the increase in (relativistic) mass with speed was a theoretical prediction based on the principles of special relativity. It wasn't an observation that required explanation. That is, it seems incongruous to try to find an alternative explanation for a prediction based on some other theory. Another thing is that the increase in (relativistic) mass is a frame of reference effect. The object itself (in its own frame of reference) does not register an increase in mass. When an object absorbs some form of energy, its rest mass increases. This does not occur if the object is simply accelerated. This is because the object in its own frame of reference is the same object regardless of its velocity relative to some other frame of reference.  14. Ah right, cool. I forgot all about relativistic and rest mass.  15. Originally Posted by KJW Actually, ... ALWAYS ... irrespective of the indefinite nature of the metric or its signature (yes, I've fallen into a similar trap!). This can be seen by going back to the definition of and the metric:  Divide both sides by : I dare to disagree on this minor point. The components of the 4-velocity in any specific Lorentz frame are given by  wherein v are the components of the ordinary 3-velocity and its magnitude respectively. The Minkowski norm of that 4-velocity ( inner product with itself ) then becomes, using the metric : This is how I originally learned it. Misner/Thorne/Wheeler in Gravitation ( page 166 ) agree with me in that u^2= -1 also - if you like I can scan the page for you and post it here.

Interestingly, I can also see where you are coming from, and I can't see anything wrong in your own argument ( except perhaps that you rely on the assumption that ds <> 0 ) - so what's the answer to this ??

EDIT : I found the exact same derivation as above in MTW on page 54 also...so who's right ?  16. Originally Posted by Markus Hanke I dare to disagree on this minor point.

...so who's right ?
Oh, I know exactly where the disagreement occurs. After all, I did say that I have fallen into a similar trap before. In this case, the most important issue isn't who's right, but why there is a difference. The disagreement arises because we are using a different definition of . My definition comes from Riemann geometry, yours comes from relativity. For my definition, a world-line is parametised by . What is depends on the signature of the indefinite metric. If the signature is , then will correspond to proper time (apart from the scaling). On the other hand, if the signature is , then will be imaginary values along the world-line of a time-like trajectory. But in either case, because this result was obtained without reference to the signature (the signature and the parametisation cancel each other in effect). By contrast, for your definition, a world-line is parametised by proper time itself. And because MTW use the signature, a time-like vector will have a square-magnitude of . But in this case, whereas proper time is real and increasing, the arc-length parameter is imaginary. This is one reason I prefer the signature, so that there is agreement between arc-length and proper time for a time-like trajectory. It should be noted that there is disagreement in the conventions used by different authors, so one should be careful not to mix up these different conventions when using different sources. The signature of the metric used in relativity is a convention, and therefore there is not a right or wrong signature. But one does need to be aware of the signature one is using.  17. I don't read a lot of texts about the mathematics of GR (it is way over my head) but almost every time I do, there seems to be a statement along the lines of "where the signature is ..." So I think people are aware of the issue.  18. Originally Posted by KJW In this case, the most important issue isn't who's right, but why there is a difference.
Agreed And because MTW use the signature, a time-like vector will have a square-magnitude of .
Yes, that's exactly it. I learned most of this stuff exclusively from GR textbooks as opposed to maths texts on DG; my very first textbook ( Prof T. Fliessbach, Allgemeine Relativitätstheorie ) also followed MTW's conventions.

It should be noted that there is disagreement in the conventions used by different authors, so one should be careful not to mix up these different conventions when using different sources.
You are right, and this is a perfect example for what you are saying.

This is one reason I prefer the signature, so that there is agreement between arc-length and proper time for a time-like trajectory.
Makes sense, I must admit. So as not to confuse myself I'll nonetheless stick with the MTW convention for now, simply so that all my personal notes are in agreement, or else I'll have a mess on my hands.
In any case, in the original issue, the main point was that the squared norm is a constant, which is something we both agree on.  Bookmarks
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