# Is there a connection between the energy distribution and time dilation?

Can anyone please help me understand what is descibed bellow?

Scenario 1. We have a pair of atomic clocks. Let's call them clock A and clock B. We switch both of them on at the same time. Clock A will stay on Earth and clock B will go with the astronauts.

Astronauts with the clock B will accelerate in direction away from the Sun for 10 years (from the astronauts' perspective) at 1 g. Than they will start braking process that will take 1 year at 10 gs. After the braking process is finished, the astronauts are not moving away from Earth anymore. They turn around and head back to Earth. The travel back will be according to the same scenario. 10 years of acceleration at 1 g and braking 1 year at 10 g.

And the astronauts are (back) home on earth.

The time on clock B is 22 years. What time is on clock A?

Scenario 2 In opposite scenario (1 year of acceleration with 10 gs and 10 year of braking with 1 g, there and back) the astronauts will "travel to future" as the clock B shows 22 years while clock A 372...

So to ask more general question: If astronauts with clok B in scenario 1&2 always spend 22 years, use same energy and reach same maximum speed...But trave distance will be different. Will it have impact on time dilation? Does the way the energy is being distributed has any impact on the time dilation?

## 2 Answers

If astronauts with clok B in scenario 1&2 always spend 22 years, use same energy and reach same maximum speed...But trave distance will be different. Will it have impact on time dilation?

Yes. The proper time experienced by the clock is given by $$\int ds$$, where $$ds^2=dt^2-dx^2-dy^2-dz^2$$ (in units with $$c=1$$). The proper time depends on the detailed shape of the clock's world-line, not just on some parameter like maximum energy.

• Thank you Ben. Could you please take a look at my comments in conversation with octonion bellow. Under his answer? I would appreciate elaboration. Thank you in advance – Miroslav Řešetka Jan 17 at 14:53

Actually, although it wasn't obvious to me at first, I think your intuition is right. In this special case of constant acceleration followed by constant braking, If you keep the maximum kinetic energy and total proper time the same in both scenarios, the total time elapsed on Earth will be the same too.

To see this, note that the situation where the astronaut always feels the same acceleration $$a$$ is known as the case of constant proper acceleration. If you search this term you can find a derivation of the kinematic formulas involved, but here I will simply state them so as to focus on the argument at hand.

The time $$t$$ observed by the person on Earth is related to the proper time $$\tau$$ felt by the astronaut as, $$t=\frac{1}{a}sinh(a\tau).$$ And the kinetic energy is, $$KE=m\left( cosh(a\tau)-1\right).$$

Here I am using units where $$c=1$$ for convenience.

Now if the mass $$m$$ and max $$KE$$ is constant, from the second equation, the combination $$a\tau = cosh^{-1}\left(\frac{KE}{m}+1\right)$$ is constant too. As a sidenote, this constant is the same thing as the rapidity, if you are familiar with that.

This means that the combination $$a t= sinh(a\tau)$$ is constant too. So then the ratio $$\frac{t}{\tau}=\frac{at}{a\tau}$$ is a constant that only depends on the max velocity (or the max kinetic energy if the mass is constant).

Thus even though the total time of the trip is broken up into four portions with different constant accelerations, the ratio of each portion to its proper time is the same. So no matter when you decide to switch from braking to accelerating, the total Earth time will just be this constant ratio times the total proper time. So your intuition that the Earth time will not depend on the scenario is correct in this very special case.

If you have a more complicated flight path with non constant acceleration, this factor relating proper time to Earth time will no longer be constant, and so Ben Crowell's answer is correct in general.

• Comments are not for extended discussion; this conversation has been moved to chat. – rob Jan 18 at 14:59