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I'm talking about the last few lines in the image. It says : $t_A$ (time measured by A1 as seen by A1) = $t’_A$ (time measured by A1 as seen by Bob)

Why are these two times equal?

The reason given is: "They must agree on the watch settings at the events even though they may disagree in the amount of time that passed between them".

I think, by 'watch settings', they mean the time displayed on the watch.

So, say, an event E occurs and Alice's watch displays time $t_1$ at the start of the event E (in Alice's frame), and $t_2$ at the end of the event. So Alice measures the interval to be $t_2-t_1$. Suppose Bob was also observing Alice's watch. If they both agree on the 'watch settings' at the start and end of the event, then Bob finds that Alice's watch was again showing times $t_1$ and $t_2$ at the start and end of the the event E respectively. So he also concludes that the interval observed by Alice is $t_2-t_1$.

Why would they both agree on the watch settings though? This is against 'relativity of simultaneity'. If 'Alice's watch showing $t_1$' and 'The beginning of the event E' occur simultaneously in Alice's frame, then it does not mean that both will occur simultaneously in Bob's frame as well.

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  • $\begingroup$ What does "the start of an event" mean? Events are points. $\endgroup$
    – WillO
    Aug 6, 2019 at 5:43
  • $\begingroup$ @WillO I actually meant two events, one which marks the start of time measurement and another which marks the end of it. $\endgroup$
    – Ryder Rude
    Aug 6, 2019 at 5:52
  • $\begingroup$ "I think, by 'watch settings', they mean the time displayed on the watch." Good question. It's possible to calculate the time intervals based on the velocity and distance, but unless the two watches have been synchronized, only Alice and Andy know what time their respective watches read. $\endgroup$ Mar 13, 2020 at 20:13

2 Answers 2

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Consider the definition of length. We may define it as the difference of the spatial coordinates between two points with respect to a reference frame, when it's temporal coordinates are the same. Basically length is the distance between two points when they are at the same time coordinate.

Thus the relationship between length with respect to two different reference frame may be obtained using the Lorentz Transformation for the spacial coordinates.

Consider, the following scenario, two reference frames A and B. A is stationary, while B is moving with a constant velocity v with respect to A. We assign x and t to the reference frame A, x' and t' to reference frame B.

Note I will be taking $c=1$ to the end.

Now consider at some t', we measure a length l with respect to B. The length will be equal to

$$l=(x'_p-x'_o)$$

Now let's write down the transformation equations for $x'_p$ and $x'_o$.

$$x'_p=\frac{x_p-vt}{\sqrt{1-v^2}}$$

And

$$x'_o=\frac{x_o-vt}{\sqrt{1-v^2}}$$

Now we know how x transforms and hence we will substitute them into our transformation equations.

Here I will consider a simple example, however it can be generalised further. For simplicity, let's consider measurements made at t'=0, between the points x'=0, and x'.

$$x'=\frac{x-vt}{\sqrt{1-v^2}}$$

Now as we have taken $t'=0$, we may imply that $t=vx$ and from here we can rewrite our transformation equation as

$$x'=\frac{x-v^2x}{\sqrt{1-v^2}}$$

This is going to give us

$$x'=x\sqrt{1-v^2}$$

Adding the speed of light in we get

$$x'=x\sqrt{1-v^2/c^2}$$

Now by our definition what is x'? Length as measured by the moving frame, what is x? Length as measured by the stationary frame. The key here is in the definition of length and time, which brings in the concept of length contraction, and time dilation

A more clear explanation is given by Leonard Susskind, in The Theoretical Minimum lecture series.

Also here is a special relativity space time graph, which works on the principles of Lorentz transformations, which will physically show how the effects take place

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  • $\begingroup$ I think the previous answer I have is not justifiable, I skipped most of the math, and asked you to consider referring, now I have instead worked out a proper method, and the references are still there just in case you wish to see. $\endgroup$
    – SK Dash
    Mar 13, 2020 at 18:11
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The page your refer to is taken from lecture notes of Dr. Gary Oas [1], and can be found in this link: https://web.stanford.edu/~oas/SI/SRGR/notes/SRGRLect2_2010.pdf

The last few lines in the image is a side note about the equivalence of inertial reference frames, rather than part of the proof of length contraction. I understand it in the following way. Two events happen:

[A] Andy looks at Bob (crucially, I mean here that she looks at his clock [2]). At the moment of "Bob passes Andy" Andy says "oh, it took Bob X seconds, as measured by his clock, to pass between Alice and me".

[B] Bob looks at himself (at his clock). At the moment of "Bob passes Andy" Bob says "oh, it took me X seconds to pass between Alice and Andy".

The key is "Andy looks at Bob" instead of "Andy looks at Alice". If Andy would look at Alice (or at herself, at her clock), it would be "less than X" seconds for Bob to pass between them. But Alice does not compare two different clocks in this example, we are only looking at Bob's clock.

Another way of looking at it is to say: the length between Alice and Andy is similarly contracted from both points of view (that of Bob and that of Andy), while the relative velocity between them is the same (from both points of view), and so the timing for Bob to pass the contracted length is the same for both Andy and Bob.

We are done. Let's just connect our findings to the way it is phrased in the image. $\Delta t'_B$, time measured by Bob as seen by Bob, is equivalent to event [B]. $\Delta t_B$, time measured by Bob as seen by Alice, involves Alice, so we rephrase event A:

[A'] Alice looks at Bob. At the moment of "Bob passes Andy" Alice says "oh, it took Bob X seconds to pass between me and Andy".

The time period X is the same in events [A] and [A'] since Andy and Alice are stationary in respect to each another. I could have written [A'] from the beginning, saving us a step. I just found it more comfortable to imagine the scene from Andy's perspective first.

Since it is the same X seconds in event [A'] and in event [B], we can write $\Delta t_B$=$\Delta t'_B$.

References and notes:

[1] I thank Dr. Gary Oas for his helpful explanation over email. The credit is his. Any mistakes here, however, are mine.

[2] It can be a regular clock or, as I sometimes prefer, a light clock. See for example figure 2 in the article of Fred Behroozi, "A Simple Derivation of Time Dilation and Length Contraction in Special Relativity", The Physics Teacher 52, 410 (2014). https://aapt.scitation.org/doi/10.1119/1.4895356

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  • $\begingroup$ I understand that $ t_B =t'_B$. But how can $t_A =t'_A$? For time measured by Alice as seen by Alice: Alice's clock reads 0 when Bob passes Alice, and, say, reads $t_A$ when Bob passes Andy. For time measured by Alice as seen by Bob: Bob again sees Alice's clock to read zero when he passes Alice. But when he passes Andy, the event "Alice's watch showing t_A" is not simultaneous to the event "Bob passed Andy". So when Bob looks at Alice's watch when he passes Andy, he observes Alice's watch to show a time $t'_A$ which is smaller than $t_A$ $\endgroup$
    – Ryder Rude
    May 19, 2020 at 1:42

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