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Shouldn't time speed up at high speeds because things would interact more quickly?

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- #1

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Shouldn't time speed up at high speeds because things would interact more quickly?

- #2

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If you search this forum or google the internet for "light clock" you will find the simplest explanation of why a clock that is at rest will tick more rapidly than a clock that is moving relative to it.Shouldn't time speed up at high speeds because things would interact more quickly?

- #3

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Shouldn't time speed up at high speeds because things would interact more quickly?

They don't interact more quickly in the reference frame moving at high speed. If you are in a rocket ship moving at, say, half the speed of light with respect to the earth, everything inside the ship would appear to be the same as if the ship were standing still with respect to the earth. There's no experiment you can do to distinguish the two situations.

- #4

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-- until the traveling ship returns the 'stationary' observer, at which point you can compare the ship clock to the stationary one, and they will show a difference.There's no experiment you can do to distinguish the two situations.

- #5

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- #6

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The usual way that people talk about time dilation is in terms of proper time, [itex]\tau[/itex] being velocity-dependent, [itex]\delta \tau(v) = \sqrt{1-\frac{v^2}{c^2}} \delta t[/itex].

I think that that's maybe a backwards way of looking at it. Proper time is the primary quantity. It's the quantity that has physical significance. Every time-dependent physical process is governed by proper time [itex]\tau[/itex], not coordinate time [itex]t[/itex]. [itex]\tau[/itex] determines the ticking of clocks and the aging of humans and the decay of uranium atoms. In contrast, [itex]t[/itex] is a label we place on events to locate them in spacetime. It has no direct physical significance. So in light of that, our equation for proper time should be inverted, to write:

[itex]\delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \delta \tau[/itex]

Of course, this is mathematically equivalent, but it switches what's considered an independent variable and what's considered a dependent variable. This alternative formula should be compared with the formula for constant-velocity motion:

[itex]\delta x = V^x \delta \tau[/itex]

Your position after time [itex]\delta \tau[/itex] changes by [itex]V^x \delta \tau[/itex], where [itex]V^x[/itex] is your velocity in the x-direction. In a similar way, the inverted time dilation formula

[itex]\delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \delta \tau[/itex]

can be interpreted as constant-velocity motion in the [itex]t[/itex]-direction, where your velocity [itex]V^t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

This is a little bit confusing, because the word "time" is being used in two different senses: (2) [itex]t[/itex] is the time-coordinate, which is one of the four coordinates you use to pinpoint a spot in 4-dimensional spacetime: [itex](x,y,z,t)[/itex]. (3) [itex]\tau[/itex] is proper time, which is the time that is important for any unfolding process such as aging, or radioactive decay or the advancing of a clock.

So in this way of viewing things, time dilation doesn't mean that "time slows down for moving clocks". Instead, it means "the amount of proper time that it takes to get to a given event (point in spacetime) depends on your velocity (both through the [itex]t[/itex] dimension, and the spatial dimensions)". It's not too strange that if two people are traveling toward the same destination, then the one with the largest velocity will get there sooner (as measured by the proper time spent traveling). The new idea in Special Relativity is to think of "destination" as being a point in 4-dimensional spacetime, rather than a point in 3-dimensional space.

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- #8

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No, as was clearly stated already, locally there is no change in the rate of things. Time ALWAYS flows locally at one second per second. Now, if you want to compare the clocks in the airplane and the car you WILL see a difference but the people in each will see the clocks ticking at one second per second.

- #9

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No, as was clearly stated already, locally there is no change in the rate of things. Time ALWAYS flows locally at one second per second. Now, if you want to compare the clocks in the airplane and the car you WILL see a difference but the people in each will see the clocks ticking at one second per second.

My question about being on an airplane was a rhetorical one for the OP, not a question for forum members.

- #10

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Oops. Had I been paying attention I would have realized that. Sorry.My question about being on an airplane was a rhetorical one for the OP, not a question for forum members.

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