Deformation of the space time continuum at speed = c

[petrushka.googol]

At velocity equal to c (if we visualize movement through space as the X axis and movement through time as the Y axis), time stands still.

 

But how? After all the time dimension cannot be decoupled from the space dimension and some ripples of gravitational waves could produce effects on the time dimension, as a collateral.

 

Altough the first postulate seems correct (and is thought to be) the latter seems intuitive.

 

Could anyone please shed light on this issue?

 

 

[ajb]

At velocity equal to c (if we visualize movement through space as the X axis and movement through time as the Y axis), time stands still.

This is a very simplistic view. You cannot think about proper time when moving at the speed of light. This is the source of confusion. You should not directly equate the time coordinate with the proper time as measured by an observer.

[swansont]

Nothing with mass can travel at v = c. It's not a valid reference frame.

[petrushka.googol]

Nothing with mass can travel at v = c. It's not a valid reference frame.

 

What about v~c (say 0.99999 c). At this velocity is it simplistic to assume that an object distorting space (by translation) would also warp time as a collateral? What my question is in essence is, is that at velocities approximately equal to c does time dimension get simultaneously warped and is a two dimensional model as described in the problem statement too simplistic and unrealistic. Could time be thought of as an extensible and deformable dimension of a 4-dimensional space-time where time is not just 1-dimensional but an attribute of a 4-dimensional co-ordinate axis. (x,y,z,t).

[Strange]

You can think of it as swapping motion through time for motion through space. The Lorentz transform is actually equivalent to a rotation between the time and space dimensions.

[swansont]

 

What about v~c (say 0.99999 c). At this velocity is it simplistic to assume that an object distorting space (by translation) would also warp time as a collateral? What my question is in essence is, is that at velocities approximately equal to c does time dimension get simultaneously warped and is a two dimensional model as described in the problem statement too simplistic and unrealistic. Could time be thought of as an extensible and deformable dimension of a 4-dimensional space-time where time is not just 1-dimensional but an attribute of a 4-dimensional co-ordinate axis. (x,y,z,t).

 

 

Time IS an attribute of a 4-dimensional system. We call it spacetime.

[Nicholas Kang]

Nothing with mass can travel at v = c. It's not a valid reference frame.

 

Because it involves division by zero, which is not a valid mathematics.

 

What about v~c (say 0.99999 c). At this velocity is it simplistic to assume that an object distorting space (by translation) would also warp time as a collateral? What my question is in essence is, is that at velocities approximately equal to c does time dimension get simultaneously warped and is a two dimensional model as described in the problem statement too simplistic and unrealistic

 

Time runs at half of the normal rate when an object approach 76% of the speed of light, c, iirc. Space is warped and thus distance is shorten by half when an object approach approximately 76% of c, according to Lorentz Transform.

[DimaMazin]

 

Because it involves division by zero, which is not a valid mathematics.

 

 

Time runs at half of the normal rate when an object approach 76% of the speed of light, c, iirc. Space is warped and thus distance is shorten by half when an object approach approximately 76% of c, according to Lorentz Transform.

0.76 c = 227,842,268.08 m/s

Recalculate gamma factor.

 

v=c(gamma²-1)^1/2 / gamma

in your case gamma=2 then v=259,627,884.49 m/s

Edited August 18, 2014 by DimaMazin

[xyzt]

 

Because it involves division by zero, which is not a valid mathematics.

 

 

Time runs at half of the normal rate when an object approach 76% of the speed of light, c, iirc. Space is warped and thus distance is shorten by half when an object approach approximately 76% of c, according to Lorentz Transform.

[math]L'=L \sqrt{1-(v/c)^2}[/math]

 

When [math]v=0.76c[/math],

 

[math]\sqrt{1-(v/c)^2}=0.66[/math]

Edited August 18, 2014 by xyzt

[Maxila]

0.76 c = 227,842,268.08 m/s

Recalculate gamma factor.

 

v=c(gamma²-1)^1/2 / gamma

in your case gamma=2 then v=259,627,884.49 m/s

 

Edit: *Correction to sentence below, In re-reading I saw you were correcting his .76c, sorry. I'll leave the workout in case it's helpful to him.

 

 

*You made an error < (*my reading error) somewhere if γ = 2 than speed is excatly 86.60254038c or ≈ 259627885 m/s

 

γ = (1 – v² / c² ) ^-1/2

 

γ = (1 – 67406638669573225 / 89875517873681764) ^-1/2

 

γ = (1 – .75) ^-1/2

 

γ = .25 ^-1/2

 

γ = 1 / √ .25

 

γ = 1 / .5

 

γ = 2

Edited August 19, 2014 by Maxila

[DimaMazin]

 

Edit: *Correction to sentence below, In re-reading I saw you were correcting his .76c, sorry. I'll leave the workout in case it's helpful to him.

 

 

*You made an error < (*my reading error) somewhere if γ = 2 than speed is excatly 86.60254038c or ≈ 259627885 m/s

 

 

 

259627885 ~ 259627884.49 Isn't it?

[Maxila]

259627885 ~ 259627884.49 Isn't it?

 

Yes, your's is more accurate I just rounded it. It would be exactly 259627884.4956445404

Edited August 19, 2014 by Maxila

[swansont]

 

Yes, your's is more accurate I just rounded it. It would be exactly 259627884.4956445404

 

No, you're both wrong, in a sense. If the speed is only specified to two digits, there's no point to displaying 18 digits of a calculation. In fact, since the formula calls for division by c, there's no point to calculating the speed and then dividing by c; that alone is likely to introduce errors (irrelevant errors, most likely, so maybe better to call them discrepancies). Just use the .76 value and square it, canceling the c's.

[Maxila]

 

No, you're both wrong, in a sense. If the speed is only specified to two digits, there's no point to displaying 18 digits of a calculation. In fact, since the formula calls for division by c, there's no point to calculating the speed and then dividing by c; that alone is likely to introduce errors (irrelevant errors, most likely, so maybe better to call them discrepancies). Just use the .76 value and square it, canceling the c's.

 

Making c = 1 is how I'd usually do it (essentially what you just showed); however for some people seeing c in numbers they are familiar with (i.e. 299792458 m/s) is helpful; that's why I showed it that way...... And you are right I did make errors using all those digits that I had to correct, lol

Edited August 19, 2014 by Maxila

[swansont]

 

Making c = 1 is how I'd usually do it

 

That's effectively the same thing. The larger issue is there are many physics problems where substituting numbers in too early is simply the wrong approach; it makes for more work and increased chance of error.

[Maxila]

 

That's effectively the same thing. The larger issue is there are many physics problems where substituting numbers in too early is simply the wrong approach; it makes for more work and increased chance of error.

 

I agree, I acknowledged that in my edit and noted I did make errors with all those digits that I had to correct before I posted that.

[xyzt]

 

No, you're both wrong, in a sense. If the speed is only specified to two digits, there's no point to displaying 18 digits of a calculation. In fact, since the formula calls for division by c, there's no point to calculating the speed and then dividing by c; that alone is likely to introduce errors (irrelevant errors, most likely, so maybe better to call them discrepancies). Just use the .76 value and square it, canceling the c's.

yep

[anthonycbush81]

Suppose zero could be oblonged in a funneled loss field. It would show 3dimensional fields yet zero would reallocate to center.

[xyzt]

Suppose zero could be oblonged in a funneled loss field. It would show 3dimensional fields yet zero would reallocate to center.

Your posts are total nonsense.

[Nicholas Kang]

Ah?! Who is correct? Thanks for all of your calculations but now what is the correct answer. I use my calculator to calculate the correct value and it is something like 86% of c is half the value for time dilation. Am I correct? And for swansont, I don`t understand what do you mean by cancel out the c, can you show me an example?

And what is gamma factor?

swansont, what do you mean by the speed is specified to 2 digits? What/which 2 digits?

Edited September 10, 2014 by Nicholas Kang

[swansont]

The formula has v/c in it. So if something is traveling at 0.86c then the term v/c = 0.86. (and with 0.86 these are the two digits — the speed is not specified any more precisely than that) You don't need to multiply by the actual number c on your calculator, and then divide by it, especially because you are also squaring it at some point. Do the algebra first. c/c = 1

[Nicholas Kang]

Is this calculation correct?

The formula has v/c in it. So if something is traveling at 0.86c then the term v/c = 0.86. (and with 0.86 these are the two digits — the speed is not specified any more precisely than that) You don't need to multiply by the actual number c on your calculator, and then divide by it, especially because you are also squaring it at some point. Do the algebra first. c/c = 1

 

Is my calculation showing both your statements, Dr. swansont?

Edited September 10, 2014 by Nicholas Kang

[swansont]

Right. You never needed to plug in the value of c to get the answer.

[Nicholas Kang]

Bingo! Thanks.