- #1

- 21

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter rocky14159
- Start date

- #1

- 21

- 0

- #2

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

[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.

[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.

Last edited:

- #3

- 21

- 0

- #4

- 1,997

- 5

In special relativity it is also F = m

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.

- #5

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

[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

Science Advisor

Homework Helper

- 4,786

- 22

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

- 21

- 0

- #8

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

8

- #9

- 21

- 0

- #10

- 1,997

- 5

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.8^{th}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?

Why anyone would discourage, or worse put down, the young and enthusiastic people is beyond me.

- #11

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

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

By the way, you titled this "F= ma at light speed". Since nothing

- #12

- 21

- 0

- #13

- 5,820

- 1,123

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.

...and use the Minkowski metric.As you can see it is pretty straightforward, we simply have to add the extra direction in relativity.

Last edited:

- #14

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

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.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 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.Why anyone would discourage, or worse put down, the young and enthusiastic people is beyond me.

- #15

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

Pretty standard 8...and use the Minkowski metric.

- #16

- 5,820

- 1,123

Pretty standard 8robphy said:...and use the Minkowski metric.^{th}grade material

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).

Last edited by a moderator:

- #17

- 1,997

- 5

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?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.

- #18

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

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).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?

Anyway, we're getting off topic. Are you still around

- #19

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

Certainly not. An object with rest mass m

[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

Science Advisor

Homework Helper

- 4,786

- 22

Pretty standard 8^{th}grade material

I learned that stuff in kindergarden I had P.A.M Dirac as teacher ;)

- #21

- 21

- 0

- #22

- 15

- 0

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

- 21

- 0

- #24

- 21

- 0

Hallsofivy, is there a name for the formula which you have provided me? thanks

- #25

malawi_glenn

Science Advisor

Homework Helper

- 4,786

- 22

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.

Share:

- Replies
- 3

- Views
- 2K