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About LinkBackscan someone explain how lorentz transformation works, without math?
if i have 2 results, how do i use it to calculate the difference in speed?
easy please.
January 31st, 2013, 09:07 PM
can someone explain how lorentz transformation works, without math?
if i have 2 results, how do i use it to calculate the difference in speed?
easy please.
No.Originally Posted by curious mind
what about, if i use einstein's train thought experement, and M' says the 2nd lightning occured 0.5 secs after the 1st, where M said both occured at the same time?
what i do now?
I'm not following you, here... Are you saying you want to solve problems without doing the math?
if light travels at c, and M' observed lightning 2 0.5 secs later, he can't be traveling at 1/2 c. but how can i know M's velocity?
I see the question but you didn't answer mine- are you saying you want to be taught a solution that doesn't involve math?
i want to know how that graph works.
I'm going to bow out because I think a stronger mind than my own may be much more helpful to youI have no idea how to describe what you're asking about- specifically how Lorentz Transformations work and apply, without math. Even with the math, I'd be hard pressed.
you have M at the embarkment seeing both lightning at once ... how would that graph look like?
then you have M' passing, seeing the 2nd lightning 0,5 secs later ... how would that graph look like?
then how you apply both to show M'speed?
the easiest thing to do would assume both lightnings hit M after 1 sec, but what now?
If you can't fathom the math, you can't understand the concepts.
it must be something like Ms graph will be t=1, and M' t=1.5, for the second lightning. i just want to know how that graph would look like for M.
oops
basically lorents transformations show the changes of spacetime on an event apparent to an observer. i found lorentz to be quite a good read
i know the meaning and what it's used for, i just don't know how to use it myself lol.
Seems like no one know about this method here.
You can't use Lorentz transformations without knowing the maths, that makes no sense. In fairness now, the maths needed are just normal standard algebra, there is not much to it.
It is not clear what you want to calculate but if you want to add two velocities; for example, a space ship travelling at velocity u (realtive to Earth) fires a missile at velocity v (realtive to the spaceship), what is the velocity the missile relative to Earth (s)?
oh i'm passed this. since everyone talked about LT in the other thread, i thought that graph was lorentz also. but googling lorentz gave no result.but then, 2 days ago, satc said it's minkowski; so i'm good now.
Lorentz transforms can involve generally simple math. The principle is to switch from one reference frame to another. For instance a very simple example would be: if you're measuring speed of a tennis ball within a reference frame on Earth where it is traveling at 65 miles an hour just in one direction, then inside a car (a different reference frame) traveling at 60 miles an hour in exactly the same direction, the same tennis ball would be moving at 5 mile per hour toward the car's front windshield. That's all there is to it concerning the simplest motion and calculation.
Lorentz used these calculations, among other things, to supposedly determine the amount of dimensional change in the famous Michelson-Morley experimental hardware to show how/why the equipment would be unable to measure an aether because of dimensional changes within the equipment because of its rotating motion.
The related equation for the tennis ball scenario would be x ′ = g (x – vt′) , where g (called gamma) and t ′(changed time) in this case are very close to 1 so are inconsequential, And where /x ′ is the new "x" coordinate, and x is the original 'x' coordinate, and v is velocity. Solving for v one gets 65 minus 60 equals 5 mph toward the front windshieldLT's were created for high velocity analysis where the gamma and time factors would not be one/inconsequential, unlike my simple example.
Lorentz transformation - Wikipedia, the free encyclopedia
Last edited by forrest noble; February 12th, 2013 at 08:12 PM.
Strictly speaking, that is a Galilean transform. But at these speeds they are pretty much identical.
i wasn't looking for LT, i was looking for minkowski. but i didn't know that it was minkowski, because everyone talked about LT in the other thread.
but i happened to find out how fast and easy the lorentz formulas are, after taking a lot longer getting the same result without it.
Minkowski spacetime relates to the inclusion of the extra dimension of time in a Cartesian coordinated system creating the concept of spacetime. There are no particular equations to this spacetime idea. It is a mathematical coordinate system. I think what you were inquiring about was Lorenz Transforms.
Last edited by forrest noble; February 12th, 2013 at 08:05 PM. Reason: added: this "spacetime" idea
If it is a mathematical system, how can there be "no particular equations"? It is a graphical representation of the hyperbolic rotation defined by the Lorentz transform. (As, of course, you know having a degree in mathematics.)
hmm ok, if i use the train example where the train moves at 0.7c and M' runs from one end to the other at 0.3c; then M would measure the speed of M' to be 0.33c?
but if M' where a plane flying over the train at 0.3c, relative to the train, M would measure M' about 0.83c?
I'm not sure whether this will be of any help to you, but this diagram is to shew how the Lorentz transforms play out.
The first diagram shews time dilation: where a time of 8 units in the static frame is dilated by the Lorentz factor γ = 1.25 at 0.6c.
In Einsteins description it its the increase in time, the slowing, of the 'ticks' of a clock
The second diagram shews the length contraction; Einstein's 1 metre rod in the moving frame being contracted to 0.8 metres in the static frame.
ok that is the graph for the formula i used in the other train where it was y = 1.40 for 0.7c. and time dilation = length contraction.
but on the example up i used
but did v-u for M' running inside the train, since i figured he's inside the moving system. is that correct?
could someone answer pls?
You don't say which way M' is running. Let's assume the same way as the train. Plugging these numbers into the equation (removing factors of c):
s = (v+u)/(1+vu)
s = (0.7 + 0.3)/(1+ 0.7*0.3)
s = 0.826446
If this is too complicated for you to work out, let a computer do it for you: (v+u)/(1+vu) where v=0.7 , u=0.3 - Wolfram|Alpha
I have no idea what that means (but you seem to have got the answer above)but if M' where a plane flying over the train at 0.3c, relative to the train, M would measure M' about 0.83c?
since in the first example M' is moving inside the train shouldn't it be v-u? because if M' stopped, he's still be carried away with the train at 0.7c.
but now the thing i have a problem with is this:
if the train were a ship moving at 0.7c, and te plane were a fly moving at 0.3c relative to the ship; wouldn't an observer on the ship say the fly moves at the speed of light but an observer on the shore will say the fly moves at approx 0.83c?
i keep thinking about this since 2 days and can't see how that could be.
Which way is M' travelling in the train? If he is travelling in the same direction as the train then M will add the velocities (as above).
If he is travelling in the opposite direction then you subtract the velocities: (v+u)/(1+vu) where v=0.7 , u=-0.3 - Wolfram|Alpha = approximately 0.5c.
Why does changing the train to a ship, and the plane to a fly make any difference?if the train were a ship moving at 0.7c, and te plane were a fly moving at 0.3c relative to the ship; wouldn't an observer on the ship say the fly moves at the speed of light but an observer on the shore will say the fly moves at approx 0.83c?
The plane/fly is travelling at 0.3c relative to the train/ship. Therefore the observer on the train/ship will see it travelling at 0.3c. that is what "relative" means.
oh dang, i should have re-read that wiki page. it says s=v+u is relative to the shore already, i thought it was relative to the ship. oh well 2 days wasted, but i learned a new formula lol.
[QUOTE If he is travelling in the same direction as the train then M will add the velocities (as above).][/QUOTE]
but why? having M' run at 0.3c has nothing to do with the train. if there was no train he'd still run at 0,3c. why would the result of an object moving inside a system be the same?
if he stops moving, unlike the plane, he's still travel at 0.7 c within the train. so he's causing a temporary time dilation in a closed system.
You said he is running in the train. So he must be moving faster than the train or he would stay at the same position in the train. So his velocity must be the train's velocity plus the speed is running down the train.
Imagine you are standing by the side of the road and bus goes past. Two kids are throwing a ball backwards and forwards the length of the bus. When the ball is thrown to the front it must be travelling faster than the bus (add the two velocities). When the ball is thrown to the back, it must be travelling slower than the bus (subtract them).
You have to be specific in relativity.
He is running a 0.3c relative to the train (or -0.3c if running against the motion of the train)
The train's speed is 0.7c relative to the track it is travelling on.
If there was no train he would be running at 0.3c relative to the track. (or -0.3c, relative to the track, if running the other way)
So running with the train his velocity is 0.3c relative to the train, 0.83c relative to the track.
Running against the motion of the train his velocity is -0.3c relative to the train, or 0.5c relative to the track.
One must be sure, when measuring speeds or velocities in relativity, just what they are relation to
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