Why is the Speed of Light Constant in Time Dilation?

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Discussion Overview

The discussion revolves around the concept of time dilation in the context of special relativity, specifically addressing why the speed of light remains constant for different observers, including those on a moving train and those stationary on a platform. Participants explore the implications of light's diagonal path as perceived by different observers and the relationship between speed, time, and distance in this framework.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants discuss the scenario of a light beam emitted from a moving train and how it appears to an observer on the platform, questioning the treatment of the light's diagonal path and its speed.
  • One participant suggests that the horizontal speed of the train should be factored into the calculation of the light's velocity vector, raising concerns about the possibility of exceeding the speed of light.
  • Another participant clarifies that the observer on the platform sees the light beam traveling straight out and back, while the observer on the train perceives it as traveling diagonally.
  • There is a discussion about the constancy of the speed of light, with some participants asserting that it cannot exceed c, as it is a postulate of relativity.
  • Participants also explore the concept of redshift and blueshift as perceived by the observer on the platform, depending on the relative motion of the train.
  • Questions are raised about the role of length contraction in the horizontal component of length, with some participants suggesting that it does not affect the calculations due to the relative motion of the observer and the train.
  • One participant proposes that length contraction can be derived from the light clock analogy once time dilation is established.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of the light's diagonal path and the implications of speed and length contraction. There is no consensus on how to reconcile these perspectives, and the discussion remains unresolved.

Contextual Notes

Limitations include the dependence on the definitions of speed and the assumptions regarding the constancy of light's speed across different frames of reference. The discussion also highlights unresolved mathematical steps in deriving the components of velocity and the implications of length contraction.

Who May Find This Useful

This discussion may be of interest to those studying special relativity, particularly in understanding time dilation, the behavior of light in different frames of reference, and the implications of relativistic effects on measurements.

cragar
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So when we derive time dilation, and when we have a train moving and we shoot a light beam from the bottom to the top, to an outside observer not on the train he will see the light leave at an angle. So my question is why don't we factor the horizontal speed of the train in order have the horizontal component of speed for the light wave . And then use the x and y component to find the magnitude of the vector.
http://en.wikipedia.org/wiki/File:Time-dilation-002.svg
this is a link to the picture i am talking about.

I realize that the speed of light is constant relative to all observers based on Maxwells equations.
 
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If I understand your question, the answer is yes we do.

The guy standing on the platform thinks the light beam travels straight out and back, a distance L with velocity c, taking time t = L/c.

The guy on the train thinks the light beam travels along the diagonal D, a distance sqrt(L2 + (vt')2), with velocity c, taking time t' = sqrt(L2 + (vt')2)/c. Squaring the latter, you get (c2 - v2)t'2 = L2 = c2t2, or t' = t/sqrt(1 - v2/c2), which is the Lorentz relation.
 
Bill_K,

Sounds about right. However, I think you got the POVs accidentally mixed up, given the stated scenario. I'm assuming you meant ...

The guy on the train thinks the light beam travels straight out and back ...

The guy standing on the platform thinks the light beam travels along the diagonal D ...​

GrayGhost
 
I don't think you guys understand what I'm asking. I don't see why the diagonal velocity is c from the person watching the train go by. the person watching the train going by sees the light leave at an angle. So to find the magnitude of that velocity vector we take the x component and the y component and then find the magnitude . If we did this we would get a speed greater than c. So my question is why don't we take this into account.
 
cragar said:
I don't think you guys understand what I'm asking. I don't see why the diagonal velocity is c from the person watching the train go by. the person watching the train going by sees the light leave at an angle. So to find the magnitude of that velocity vector we take the x component and the y component and then find the magnitude . If we did this we would get a speed greater than c. So my question is why don't we take this into account.

cragar said:
I realize that the speed of light is constant relative to all observers based on Maxwells equations.
You have answered your own question.

If "the speed of light is constant relative to all observers" then we cannot "get a speed greater than c".
 
cragar said:
I don't think you guys understand what I'm asking. I don't see why the diagonal velocity is c from the person watching the train go by. the person watching the train going by sees the light leave at an angle. So to find the magnitude of that velocity vector we take the x component and the y component and then find the magnitude .
Well, you know the light must remain directly above the emitter at all times (so that the light travels vertically in the x frame), so that means the x component must be the same as the velocity of the emitter, i.e. the velocity of the train. But how do you propose to find the y component? The normal method would be to start with the fact that relativity says the total magnitude of the velocity vector must be c, giving (train velocity)^2 + (y component)^2 = c^2, then solve for the y component. If you have some other suggestion about how to derive the y component, please explain.
 
I realize that the speed of light is constant relative to all observers based on Maxwells equations.[/QUOTE]

This is a postulate. In whichever direction, observer on the platform sees light beam travelling, it will be c. Not less, not more. That is why we get time dilation.
 
will the light appear to be red shifted or blue shifted to the observer watching the train go by.
 
cragar said:
will the light appear to be red shifted or blue shifted to the observer watching the train go by.
Blue shifted when he sees the train approaching him, red shifted when he sees it moving away from him.
 
  • #10
ok thanks , one more question about it, why don't we factor in length contraction into the horizontal component of length. Is this because the ruler on the trains also contracts so it doesn't matter . And i think that length contraction is also derived from this.
 
Last edited:
  • #11
cragar said:
ok thanks , one more question about it, why don't we factor in length contraction into the horizontal component of length.
The horizontal component is the distance (in the observer's frame) the top and bottom mirror travel in the time t it take the light to get from one to the other, length contraction doesn't change the fact that if something is moving at v in my frame for time t, it travels a distance of v*t. Length contraction applies when you know the distance between two objects (or two ends of a single object) in their own rest frame, and you want to know the distance between them in your frame where both are moving.
cragar said:
And i think that length contraction is also derived from this.
Yes, you can derive length contraction from the light clock once you've already derived time dilation, just imagine setting the light clock on its side so that instead of one mirror being above the other, both mirrors are on the floor, the axis between them being parallel to the direction the train is moving in your frame. Then since the time to bounce from one to the other and back in your frame should be dilated by the same amount as a vertical clock (because they both tick at the same rate in the train rest frame, so they have to in your frame too), you can figure out how far apart they need to be for this to be true.
 
  • #12
thanks for your answer
 

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