# I Calculation of Time Dilation with Online Calculators

#### drkfuture

Summary
Online Time Dilation Calculator gives different answer for same input
I m a noob, I wanted to figure out the time dilation for an observer on earth who passed 50,000 years, and the Traveler at light speed(99.9%) passed ONE DAY. I mean we pass 50,000 years on earth and the Traveler spends One day.Problem is I tried many online calculator, but each one is giving me different result with the same inputs.I m confused.I want to know the result. I also want to consider the LENGTH CONTRACTION/DISTANCE DILATION while calculating this.I m not very good at maths. Can somebody figure this out for me. Please make sure ur calculation is right, bcoz each online Time Dilation calculator gives different answer,which is frustrating!!.

Considering the aforementioned scenario, I want to find out:
1-Time observed by the person on earth
2-Distance Diluted by the traveler

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#### mfb

Mentor
What did the calculators give you?
Did you plug in the time dilation factor in the formula and calculate it yourself?

(the speed will be much higher than 99.9% the speed of light)

#### Ibix

Back of the envelope, you'll need fifteen decimal places of precision to do this calculation. It's quite likely that this will get rounded off by a calculator, and it'll just tell you that the answer is the speed of light. If it doesn't, the answer will quite likely depend on the order in which the programmer wrote the maths because that will affect what gets rounded off and when.

It'll also depend on whether you think of a year as 365 days, or 365.25 days, or some more precise figure. Which number did you intend to use?

#### pervect

Staff Emeritus
If I'm understanding correctly, you want a gamma factor of $\gamma$ = 18,262,110, which is 50,000 years / 1 day, using Google calculators notion of a year as approximately 365.25 days.

The above gamma factor is the required time dilation factor, also the required length contraction factor. "Distance dilation" doesn't really make sense.

You didn't ask for the required velocity, which I make out to be
$$c \, \sqrt{1 - \frac{1}{\gamma^2}} \approx c \, \left(1 - \frac{1}{2 \, \gamma^2} \right)$$

#### DrGreg

Gold Member
You didn't ask for the required velocity, which I make out to be
$$c \, \sqrt{1 - \frac{1}{\gamma^2}} \approx c \, \left(1 - \frac{1}{2 \, \gamma^2} \right)$$
That's what I got, too, which works out to be 449 nm/s slower than the speed of light (that's half a thousandth of a millimetre per second slower).

#### drkfuture

THANK U ALL FOR THE REPLY. Actually I guess, I need to rephrase my question, I want to know if The astronaut leaves on a deep space trip traveling at 99.9% the speed of light,upon returning earth, the astronauts clock has measured TWO DAYS. How much time we, on earth, will spend ?

#### FactChecker

Gold Member
2018 Award
This is better. Your original question was over-specified and the numbers may not have been compatible. You are not free to arbitrarily specify all three of time-here, time of traveler, velocity. You can specify two values and then must calculate the third compatible value.
What you have now is a standard problem of finding the Lorentz factor, γ=1/√1−.9992=22.366γ=1/1−.9992=22.366 and using it to calculate the time-duration in our reference frame: T = 22.366 * 2 = 44.73 days.

CORRECTION(?): I did not consider the issue of acceleration for the astronaut to turn around and come back. If this question is supposed to consider that, it is a much different problem like the twins paradox. In fact, any problem where the traveler returns home is different. The answer here is for an astronaut remaining in a constant moving inertial reference frame. So he can not turn around and come home.

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#### Pencilvester

Have you tried doing the calculation yourself? Here are the relevant equations, called the Lorentz transformation: $$x’ = \gamma (x - vt)$$ $$t’ = \gamma (t - vx/c^2)$$ where $\gamma$ is the Lorentz factor which equals $1/\sqrt{1-v^2/c^2}$. This take events in one inertial coordinate system and gives you the coordinates of that event in a different inertial coordinate system.

You are concerned with comparing coordinate time of the earthbound’s (E.B.) inertial frame to the traveller’s proper time (i.e. his own elapsed time). The traveller cannot be tied to a single inertial frame for the reason that he must change directions in order to return to Earth. If we assume the traveller is maintaining constant speed relative to the Earth and that he can withstand infinite acceleration from an instantaneous change in direction, then we can stitch together two different inertial frames to attach to the traveller—an outbound one and an inbound one. Due to symmetry, we can just calculate E.B.’s coordinate time from the traveller’s departure to his point of turnaround and simply multiply by 2 to get the total time, and this allows us to use the Lorentz transformation. If you assign the unprimed coordinates to the traveller and the primed to the E.B. observer, then you can label the relevant events as such:

Event A: Traveller leaves Earth—traveller could label as $x=0 ,t=0$ and E.B. observer could label this as $x’=0 , t’=0$
Event B: Traveller makes his instantaneous turnaround to head back to Earth—this is the halfway point in the traveller’s journey, so $t$ should equal $(t_A + 2 ~ \mathrm{days})/2 = 1 ~ \mathrm{day}$, and $x=0$ since this is the traveller’s own frame of reference, and he does not consider himself moving in his own frame.

This is all you need to be able to calculate the coordinate time E.B. assigns to event B, then you just multiply by 2 to get the total time. You can use this process to plug in any velocities and elapsed times that you want, just make sure you keep track of which times belong to which frames.

#### FactChecker

Gold Member
2018 Award
Due to symmetry, we can just calculate E.B.’s coordinate time from the traveller’s departure to his point of turnaround and simply multiply by 2 to get the total time, and this allows us to use the Lorentz transformation.
This sounds wrong to me. The switch from the outbound frame to the inbound frame includes a change of time coordinates that should not be ignored.
CORRECTION: I stand corrected. When calculating the elapsed time on Earth, there is symmetry and no complications.

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#### Nugatory

Mentor
This sounds wrong to me. The switch from the outbound frame to the inbound frame includes a change of time coordinates that should not be ignored.
@pencilvester's method works because by symmetry the elapsed time in the earth frame for the outbound leg and the return leg will be the same. Thus, once we've calculated the outbound time we don't need to do another calculation for the return leg because we already know the answer will be the same as for the outbound leg.

You are right that we'd be using a different frame if we did choose to do the calculation for the return leg instead of taking advantage of the symmetry.

#### FactChecker

Gold Member
2018 Award
@pencilvester's method works because by symmetry the elapsed time in the earth frame for the outbound leg and the return leg will be the same. Thus, once we've calculated the outbound time we don't need to do another calculation for the return leg because we already know the answer will be the same as for the outbound leg.
I stand corrected. When calculating the elapsed time on Earth, there is symmetry and no complications.

#### pervect

Staff Emeritus
Pencilvester's detailed argument shows us that t' =$\gamma$ t, when we substitute x=0 into the Lorentz transform

$$t' = \gamma \left(t - \frac{vx}{c^2} \right)$$

And since $\gamma = 1 / \sqrt{1-\beta^2}$ and $\beta = v/c = .999$, at 99.9% of the speed of light we can compute that $\gamma \approx 22.37$

So the two day trip for the astronaut will take approximately 44.7 days on Earth at 99.9% of the speed of light.

#### drkfuture

This is better. Your original question was over-specified and the numbers may not have been compatible. You are not free to arbitrarily specify all three of time-here, time of traveler, velocity. You can specify two values and then must calculate the third compatible value.
What you have now is a standard problem of finding the Lorentz factor, γ=1/√1−.9992=22.366γ=1/1−.9992=22.366 and using it to calculate the time-duration in our reference frame: T = 22.366 * 2 = 44.73 days.
THANK U VERY MUCH BRO,
I dont know the LENGTH of the space traveler in my case, I also dont know how the Length Contraction affect the result of the above formula, can u explain a bit? All I need to know the relative time we , on earth, will observe for the traveler who traveled TWO DAYS in space.

One more thing can u figure out the distance of the remote planet the traveler went in one day (single trip)?

this is exactly what I got from another ONLINE CALCULATOR, which seems to be OK. Check the Screenshot
BELOW, I got the time dilation for single trip here (then I multiplied by 2).Does it look OK to you?

#### drkfuture

CORRECTION(?): I did not consider the issue of acceleration for the astronaut to turn around and come back. If this question is supposed to consider that, it is a much different problem like the twins paradox. In fact, any problem where the traveler returns home is different. The answer here is for an astronaut remaining in a constant moving inertial reference frame. So he can not turn around and come home.
This is a bit complex for me, r u telling me that this formula cant be applied for ROUND TRIP? If a traveler goes to a distant planet and comes back to earth, cant I use this formula ?

#### Ibix

This is a bit complex for me, r u telling me that this formula cant be applied for ROUND TRIP? If a traveler goes to a distant planet and comes back to earth, cant I use this formula ?
You can use it. There are some subtleties with return journeys that frequently lead people into traps, but if all you want to know is how long passes on Earth while someone does an out-and-back trip lasting two days by their clocks, this result is fine.

#### FactChecker

Gold Member
2018 Award
This is a bit complex for me, r u telling me that this formula cant be applied for ROUND TRIP? If a traveler goes to a distant planet and comes back to earth, cant I use this formula ?
You can use it for the answer you asked for. I was confused. Sorry. As @Ibix says, as long as you are only considering the proper time on Earth (how long an Earthling thinks the trip took in Earth time), there is no complication due to the turn-around of the astronaut.

On the other hand, suppose you ask about what the astronaut thinks that the Earth's elapsed time is in slow Earth time. It is wrong to say that it is shorter than two days by the factor of $1/\gamma$. The turn-around and the change of frame come into play. This issue comes up in a famous "Twin's Paradox". It is not really a paradox if the turn-around is handled correctly.

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