Questions about SR

  • #1

Main Question or Discussion Point

Two cars are stationary and next to each other and not moving relative to each other.

According to SR their clocks should be synchronized. Am I correct in saying that?

Also by saying this am I correct in saying that they agree on the time of events?

Now what would happen if one car starts going at .5c? What is his time now? Why exactly does it mean for his clocks to slow down?
 

Answers and Replies

  • #2
260
1
Two cars are stationary and next to each other and not moving relative to each other.

According to SR their clocks should be synchronized. Am I correct in saying that?
Good so far.

(**Technically you would have to synchronize the clocks in the first place, but this is implied.)

Also by saying this am I correct in saying that they agree on the time of events?
Yes, this is correct.

Now what would happen if one car starts going at .5c? What is his time now? Why exactly does it mean for his clocks to slow down?
If the cars are moving apart at 0.5c, then both cars will observe each others' clocks running more slowly than their own. This means that, for every tick of one cars clock, the other car's clock would appear to take 1.1547 times to tick once. (At 0.5c, γ=1.15470).
 
  • #3
49
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^how did you know at .5c y=1.15470? Did you manually put the numbers in and figure it out, or is there something you referenced that maybe you could link?
 
  • #4
260
1
^how did you know at .5c y=1.15470? Did you manually put the numbers in and figure it out, or is there something you referenced that maybe you could link?
I did [itex]1/ \sqrt{1-0.5^2}=1/ \sqrt{0.75}[/itex] on my calculator, though I did just find this: http://www.wolframalpha.com/entities/calculators/lorentz_factor_formula/fl/z1/6z/ [Broken].
 
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  • #5
1,538
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Good so far.
If the cars are moving apart at 0.5c, then both cars will observe each others' clocks running more slowly than their own. This means that, for every tick of one cars clock, the other car's clock would appear to take 1.1547 times to tick once. (At 0.5c, γ=1.15470).
First we have to be clear about what we mean by observe.

When two cars move away from each other with a velocity of 0.5c they actually measure each others clocks to go at about 58% of their own clock.

They can make a calculation taking into account the transfer time of light and only then will they infer the other clock goes at 87% of their own clock. But that calculation has no direct physical significance.

But what is measured is not the Lorentz factor instead the Doppler factor is measured.
 
  • #6
ghwellsjr
Science Advisor
Gold Member
5,122
146
First we have to be clear about what we mean by observe.

When two cars move away from each other with a velocity of 0.5c they actually measure each others clocks to go at about 58% of their own clock.

They can make a calculation taking into account the transfer time of light and only then will they infer the other clock goes at 87% of their own clock. But that calculation has no direct physical significance.

But what is measured is not the Lorentz factor instead the Doppler factor is measured.
Your post leads to more questions:

Is there any physical significance, direct or indirect, to the measurement that each car makes of the rate of the other one's clock?

What calculation are you talking about that takes into account the transfer time of light?

How does each car know what that transfer time of light is? Is this another calculation, observation or measurement?

Is there any direct or indirect physical significance to that transfer time of the light?

Can each car observe, measure or calculate the velocity at which they move away from each other? If so, how?

Is there any direct or indirect physical significance to the observation, measurement or calculation of that velocity?
 
  • #7
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Is there any physical significance, direct or indirect, to the measurement that each car makes of the rate of the other one's clock?
Most definitely, any signal or form of radiation is adjusted by the same factor as the clock. So for instance if each car shines a light with a certain frequency this frequency will be measured at about 58% of the original frequency.

We calculate:
[tex]\Large {\frac {\sqrt {1-{v}^{2}}}{1+v}}[/tex]
What calculation are you talking about that takes into account the transfer time of light?
When two objects are retreating or approaching each other the total time for a light signal to reach the other object changes with time, this has a resp. red and blueshift effect. If we disable this part we have a remainder of the total Doppler shift which can be expressed by the Lorentz factor.

How does each car know what that transfer time of light is? Is this another calculation, observation or measurement?
Let me turn it around, they can measure the Doppler shift, the transfer time of light cannot be directly observed and, as a consequence, the Lorentz factor cannot be directly measured either.

However there is one interesting exception, and that is when the two cars move laterally, in this case the Doppler factor is equal to the Lorentz factor.

Is there any direct or indirect physical significance to that transfer time of the light?
Yes there is, and this is very important in case of differential aging when for instance one of the cars turns around. The 'at home' car will count a fewer number of waves.

Can each car observe, measure or calculate the velocity at which they move away from each other? If so, how?
They can if both cars emit for instance some reference frequency. Based on the actual measurement of the frequency (and for simplicity I assume longitudinal motion only) and expressing that as a factor they could calculate it as such:
[tex]\Large v = -{\frac {-1+{{\it ratio}}^{2}}{1+{{\it ratio}}^{2}}}[/tex]
Is there any direct or indirect physical significance to the observation, measurement or calculation of that velocity?
Yes there is, the fact that two objects are in relative motion is obviously of physical significance. For instance at one point things may bump into each other.
 

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