F=ma at light speed

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Im just an amateur, and I read that Newton's Equation F=ma doesnt work when approaching light speeds because of a relative increase in mass at higher velocicites. What is the reason for this, if any, and is there a ratio between veolcity and relative mass I could use to make my calculations more accurate when dealing with light speed? Thanks!
 

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  • #2
Hootenanny
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[itex]F = ma[/itex] may not work, but the full version of Newton's Second law does,

[tex]\mathbf{F} = \frac{d\mathbf{p}}{dt}[/tex]

even at velocities close to the speed of light. However, you should note that unlike the classical case, the acceleration is not generally in the same direction as the force.
 
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  • #3
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Thanks, but can you tell me what the variables d, p, and t are? Im new here and not yet familiar with most full formulas
 
  • #4
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Im just an amateur, and I read that Newton's Equation F=ma doesnt work when approaching light speeds because of a relative increase in mass at higher velocicites. What is the reason for this, if any, and is there a ratio between veolcity and relative mass I could use to make my calculations more accurate when dealing with light speed? Thanks!
In special relativity it is also F = m0A. And this forumla always works!

But the difference is that A is a 4-vector. A is the 4-acceleration vector which is basically the derivative of the 4-velocity vector with respect to proper time. And the 4-velocity vector is basically the change in direction in spacetime with respect to proper time. But note that we have four, not three, directions in spacetime!
In addition there is the 4-energy-momentum vector which is mass times the 4-velocity vector.

Four vectors are simple and elegant.

As you can see it is pretty straightforward, we simply have to add the extra direction in relativity. :smile:
 
  • #5
Hootenanny
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Thanks, but can you tell me what the variables d, p, and t are? Im new here and not yet familiar with most full formulas
p is the momentum of the particle and,

[tex]\frac{d}{dt}[/tex]

Is the first derivative with respect to time. I'm guessing that you haven't done any calculus if you haven't seen this notation before. If that is the case, a more appropriate formulation would be,

[tex]m\mathbf{a} = \mathbf{F} - \frac{\left(\mathbf{F}\mathbf{\cdot}\mathbf{v}\right)\mathbf{v}}{c^2}[/tex]

You should note that the product [itex]\left(\mathbf{F}\mathbf{\cdot}\mathbf{v}\right)[/itex] isn't simply a scalar multiplication, rather it's the dot product between the two vectors (force and velocity).
 
  • #6
malawi_glenn
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the d's mean that you take the derivative of momentum (p), with respect to time (t)

Veolcity = [tex] \frac{d\vec{r}}{dt} [/tex]

Acceleration = [tex] \frac{d\vec{v}}{dt} [/tex]

Momentum = [tex] \vec{p} = \gamma m \vec{v} [/tex]

Where: [tex] \gamma = \frac{1}{\sqrt{1-(v/c)^2}} [/tex]
 
  • #7
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You lost me. Could you explain that in terms that an 8th grader who's only in Geometry could understand?
 
  • #8
Hootenanny
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You lost me. Could you explain that in terms that an 8th grader who's only in Geometry could understand?
8th grade, so you're about fourteen right? Without wanting to sound too harsh, I'm afraid that you're going to have a difficult time understanding this. Perhaps in a few years when you've done some more advanced mathematics (and basic physics) courses you might be able to have another go, but at the moment relativity is a little above your level. Which begs the question why are you studying SR? Is it for a project, or are you just generally interested?
 
  • #9
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I am interested, and I wanted to see around how much U-235 you would have to fission (at 100% efficiency) to propel the space shuttle to light speed (i know its not that simple, but i wanted a round number to see how much you would need) and at the moment all I can do is use F=ma because thats all I know, but I was hoping that there would be some sort of ratio to how fast you go to how much mass you gain, because then I could simply tack that on to my equation and plug in the numbers and solve again, with a more accurate answer.
 
  • #10
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8th grade, so you're about fourteen right? Without wanting to sound too harsh, I'm afraid that you're going to have a difficult time understanding this. Perhaps in a few years when you've done some more advanced mathematics (and basic physics) courses you might be able to have another go, but at the moment relativity is a little above your level. Which begs the question why are you studying SR? Is it for a project, or are you just generally interested?
I think it is great that someone who is 14 years old is asking questions about relativity. I think a 14 year old should already understand derivatives. Myself I was 12 years old when I became familiar and understood special relativity.

Why anyone would discourage, or worse put down, the young and enthusiastic people is beyond me.
 
  • #11
HallsofIvy
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You lost me. Could you explain that in terms that an 8th grader who's only in Geometry could understand?

Now that's a challenge! You know of us had to spend years and years in classes just to beign to understand this stuff!

The "derivative" of a quantity is essentially its instantaneous rate of change as opposed to an "average" rate of change. The "a" (acceleration) in F= ma is the instantaneous rate of change of velocity. That could be written [itex]F= m dv/dt[/itex]. That is true as long as the mass (m) is constant. If it is changing itself it is better to say "Force = rate of change of momentum" and momentum, of course, is "mass times velocity" so "F= rate of change of mass times velocity" or F= d(mv)/dt.

By the way, you titled this "F= ma at light speed". Since nothing with mass can move at light speed, the question is vacuous.
 
  • #12
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So I take it no one has ever found the ratio between an object's velocity and its relative gained mass, am I correct in assuming this?
 
  • #13
robphy
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I think it is great that someone who is 14 years old is asking questions about relativity. I think a 14 year old should already understand derivatives. Myself I was 12 years old when I became familiar and understood special relativity.

Could you elaborate on what "understood" means in this sentence?
I was about 13 when I was introduced to relativity [via a video on PBS]... but understanding it took a while.

As you can see it is pretty straightforward, we simply have to add the extra direction in relativity. :smile:
...and use the Minkowski metric.
 
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  • #14
Hootenanny
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I am interested, and I wanted to see around how much U-235 you would have to fission (at 100% efficiency) to propel the space shuttle to light speed (i know its not that simple, but i wanted a round number to see how much you would need) and at the moment all I can do is use F=ma because that's all I know, but I was hoping that there would be some sort of ratio to how fast you go to how much mass you gain, because then I could simply tack that on to my equation and plug in the numbers and solve again, with a more accurate answer.
It's good that your interested, but there are a couple of points to mention here. Firstly, no matter how much energy you have your space shuttle will never reach the speed of light, it can approach arbitrarily close to the speed of light, but it will never actually get there. Secondly as I have said previously, the acceleration produced by a given force isn't necessarily in the same direction as the force. One can use the concept of relativistic mass, as described by MG, but you simply can't substitute this force the mass term in the classical expression.
Why anyone would discourage, or worse put down, the young and enthusiastic people is beyond me.
It is not an insult to tell someone that they haven't the sufficient knowlage to understand a concept. It is far worse to lead someone to believe that they do understand something when they don't.
 
  • #15
Hootenanny
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...and use the Minkowski metric.
Pretty standard 8th grade material :rolleyes:
 
  • #16
robphy
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robphy said:
...and use the Minkowski metric.
Pretty standard 8th grade material :rolleyes:

I was just making the point that
"relativity" is more about adding a dimension to "space" and dealing with a higher-dimensional space. Indeed, most of the trouble folks have with relativity is not about its dimensionality. Just consider every example with "motion in the x-direction".

There are ways to handle the non-euclidean-signature metric without explicitly invoking the metric or the Lorentz transformation.... but most textbooks don't do them (or seem to know about them).
 
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  • #17
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It is not an insult to tell someone that they haven't the sufficient knowlage to understand a concept. It is far worse to lead someone to believe that they do understand something when they don't.
In my generation the concept of derivatives was taught at schools when the student was between 13 and 14 years old. Is it now different?
 
  • #18
Hootenanny
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In my generation the concept of derivatives was taught at schools when the student was between 13 and 14 years old. Is it now different?
I would assume so since the OP states that he hasn't met them before. From what I know of the US education system calculus is usually introduced in the 11th grade (16-17).

Anyway, we're getting off topic. Are you still around rocky14159?
 
  • #19
HallsofIvy
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So I take it no one has ever found the ratio between an object's velocity and its relative gained mass, am I correct in assuming this?
Certainly not. An object with rest mass m0, moving at speed v, will have relativistic mass of
[tex]\frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]

If you know about mass increasing I'm suprised you hadn't see that formula before.
 
  • #20
malawi_glenn
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Pretty standard 8th grade material :rolleyes:

I learned that stuff in kindergarden :biggrin: I had P.A.M Dirac as teacher ;)
 
  • #21
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I'm here, but I am in 8th grade and I have never even heard of any of this, but, I also got most of this information from other sources than school
 
  • #22
Hmm, well a little late, I know. But Rocky, since you're in geometry I assume you've been in Algebra I. In Algebra, you learn about rise over run, you count up how many boxes a line goes and how many boxes over, it goes, divide the up by the over and you've got your slope. This is great for straight lines, but when the line curves around counting boxes is nearly impossible. It is presented this way at the beginning of calculus as a refresher, but once you learn derivatives you'll no longer need this.

dp is the derivative of the momentum (m*v), dividing that by the derivative of time and you get m*v/t. Mass(kg)*Velocity(meters/second) divided by seconds gives you mass*(acceleration (meters/seconds/seconds)), or meters/seconds^2.

I hope that helps you get your head around this stuff a little more, and continue on with the thoughts, wondering is the beginning of understanding.
 
  • #23
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HallsofIvy, thank you, the formula you have provided shows to me undoubtedly that you cannot go light speed, but very close to it, and i shall use 185999 for my velocity now, and I think i shall gain substantial knowledge and accurate answers from this.
 
  • #24
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Hallsofivy, is there a name for the formula which you have provided me? thanks
 
  • #25
malawi_glenn
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Hallsofivy, is there a name for the formula which you have provided me? thanks

I thougt it was hootenanny in post #5 ? The formula is called Newtons second law.
 
  • #26
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Not that one, the one that HallsofIvy posted in post 19
 
  • #27
malawi_glenn
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ah, it is called "relativistic mass"

[tex] m = \gamma m_0 [/tex]

Where [tex] m_0 [/tex] is the rest mass, and [tex] \gamma = 1/\sqrt{1-(v/c)^2} [/tex] is called the Lorentz factor.
 
  • #28
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By starting from the equation

[tex]
F=\frac{dp}{dt},
[/tex]

it is possible to solve a 3x3 matrix [itex]m[/itex] so that

[tex]
F=ma
[/tex]

is true for all F and |v|<c. So F=ma can be made true, if we want so :wink: At least for the massive particles. I haven't tried it for the massless ones, though.
 
  • #29
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You should note that the product [itex]\left(\mathbf{F}\mathbf{\cdot}\mathbf{v}\right)[/itex] isn't simply a scalar multiplication, rather it's the dot product between the two vectors (force and velocity).

Is this supposed to mean something? :confused:
 
  • #30
Hootenanny
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Is this supposed to mean something? :confused:
It means exactly what it says, I was merely emphasising the point since the OP was unsure as to notation.
 
  • #31
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What is the difference between scalar and dot products in this context?
 
  • #32
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What is the difference between scalar and dot products in this context?

Scalar multiplication vs the vector dot product? The latter is of course also known as the scalar product but we're talking about multiplication of two scalars vs a vector operation.
 
  • #33
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So by "scalar multiplication" hootenanny meant a multiplication of two scalars? That would explain the confusion. But the dot product is also called the scalar product.

dot product = inner product = scalar product

Those are different names for the same thing.
 
  • #34
malawi_glenn
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yeah, but he meant product by two scalars (the OP are unfamilar with linear algebra and vectors so it doesent matter)
 
  • #35
Hootenanny
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Generally scalar multiplication is taken too mean two things: either the multiplication of two scalar quantities or the multiplication of a vector by a scalar quantity. Whereas the term scalar product means the inner or dot product between two vectors.

Notice that in my post I said scalar multiplication and not scalar product, which are two different operations.
 

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